During the titration of a weak diprotic acid `(H_(2)A)` against a strong base `(NaOH)`, the pH of the solution half-way to the first equivalent point and that at the first equivalent point are given respectively by:

Options

A. `pK_(a1) and pK_(a1)+pK_(a2)`
B. `sqrt(K_(a1)C) and(pK_(a1)+pK_(a2))/2`
C. `pK_(a1)and(pK_(a1)+pK_(a2))/2`
D. `pK_(a1)andpK_(a2)`

Answer

To solve the problem of determining the pH of a weak diprotic acid (H₂A) during its titration with a strong base (NaOH), we will analyze the situation at two specific points: halfway to the first equivalent point and at the first equivalent point.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The weak diprotic acid (H₂A) reacts with NaOH to form NaHA (the conjugate base of the first dissociation) and water.
- The reaction can be represented as:
\[ \text{H}_2\text{A} + \text{NaOH} \rightarrow \text{NaHA} + \text{H}_2\text{O} \]

2. **Halfway to the First Equivalent Point**:
- At halfway to the first equivalent point, half of the H₂A has been neutralized by NaOH.
- If we start with 'A' moles of H₂A, at this point we have:
- Moles of H₂A remaining: \( \frac{A}{2} \)
- Moles of NaHA formed: \( \frac{A}{2} \)
- Since we have equal concentrations of the weak acid (NaHA) and its conjugate base (H₂A), we can use the Henderson-Hasselbalch equation for buffer solutions:
\[ \text{pH} = \text{pK}_a1 + \log\left(\frac{[\text{NaHA}]}{[\text{H}_2\text{A}]}\right) \]
- Given that the concentrations of NaHA and H₂A are equal, the log term becomes zero:
\[ \text{pH} = \text{pK}_a1 \]

3. **At the First Equivalent Point**:
- At the first equivalent point, all of the H₂A has been converted to NaHA.
- The solution now contains only the salt NaHA, which is an amphiprotic salt.
- The pH at this point can be calculated using the average of the two dissociation constants (pK₁ and pK₂):
\[ \text{pH} = \frac{\text{pK}_a1 + \text{pK}_a2}{2} \]

4. **Summary of Results**:
- The pH halfway to the first equivalent point is:
\[ \text{pH} = \text{pK}_a1 \]
- The pH at the first equivalent point is:
\[ \text{pH} = \frac{\text{pK}_a1 + \text{pK}_a2}{2} \]

### Final Answer:
- The pH halfway to the first equivalent point is \( \text{pK}_a1 \).
- The pH at the first equivalent point is \( \frac{\text{pK}_a1 + \text{pK}_a2}{2} \).